Q 51.

Question

Find a vector of length 3 that points in the direction opposite to $<1, - 2, 3>$ .

Step-by-Step Solution

Verified

Answer

The required vector is $\frac{3}{\sqrt{14}} <- 1, 2, - 3>$ .

1Step 1. Given information.

The given vector is:

$v = <1, - 2, 3>$ and its length is 3.

2Step 2. Find the norm of the vector.

$v = <1, - 2, 3> \frac{1}{||v||} = \sqrt{1^{2} + {(- 2)}^{2} + 3^{2}} = \sqrt{1 + 4 + 9} = \sqrt{14}$

Therefore, the norm of the given vector is $\sqrt{14}$ .

3Step 3. Find the vector.

We know that the vector in the direction of v is $\frac{a}{||v||} (- v)$ . Here a is the magnitude of the given vector and the negative sign shows the direction opposite to $v = <1, - 2, 3>$ .

The required vector is:

$\frac{a}{||v||} v = \frac{3}{\sqrt{14}} (- <1, - 2, 3>) = \frac{3}{\sqrt{14}} <- 1, 2, - 3>$

Therefore the vector in the direction opposite to $v = <1, - 2, 3>$ is $\frac{3}{\sqrt{14}} <- 1, 2, - 3>$ .

Q 50.

Q. 52

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